> So, is there a alternative way where I can do different 
> operation depending on the position of node(element) to be 
> even or odd.

Depends exactly what you mean by "position", but you can probably do

<xsl:if test="position() mod 2 = 0">

or

<xsl:if test="count(preceding-sibling::*) mod 2 = 0">

Michael Kay